A straight line has one gradient; a curve has a different gradient at each point. Differentiating a function gives a new function, the derivative, that outputs the gradient of the tangent at any \(x\). Six standard results, a rule for constants and sums, and the chain rule for functions inside functions are all you need for every derivative in 0606.
The derivative \(\dfrac{dy}{dx}\) or \(f'(x)\) gives the gradient of the tangent to \(y = f(x)\) at each point. Learn these, with \(x\) in radians: \(x^{n} \to nx^{n-1}\) (any rational \(n\)), \(\sin x \to \cos x\), \(\cos x \to -\sin x\), \(\tan x \to \sec^{2} x\), \(e^{x} \to e^{x}\), \(\ln x \to \dfrac{1}{x}\). Constant multiples stay and sums are differentiated term by term. For a function of a function, the chain rule says \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\): differentiate the outside, keep the inside, then multiply by the derivative of the inside.
Fig. 1: the curve y = x² with P(1, 1). The chord to Q₁(2, 4) has gradient 3 and the chord to Q₂(1.5, 2.25) has gradient 2.5. As Q moves along the curve towards P the chord gradients 3, 2.5, 2.1, 2.01, ... close in on 2, the gradient of the tangent at P
Section 01
The gradient of a curve.
The gradient of a curve at a point is the gradient of the tangent there: the straight line that just touches the curve at that point. In Fig. 1, chords from \(P(1, 1)\) on \(y = x^{2}\) to nearer and nearer points Q have gradients 3, 2.5, 2.1, 2.01, closing in on 2, so the gradient at \(P\) is 2. Doing this at every \(x\) gives the derived function, or derivative. For \(y = x^{2}\) it is \(2x\). You will not be asked to do the limit yourself; you use the rules below.
Notation
Meaning
\(\dfrac{dy}{dx}\) or \(f'(x)\)
first derivative: the gradient of \(y = f(x)\)
\(\dfrac{d^{2}y}{dx^{2}}\) or \(f''(x)\)
second derivative: differentiate again, \(\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)\)
The power rule: \(\dfrac{d}{dx}\left(ax^{n}\right) = anx^{n-1}\) for any rational \(n\): multiply by the power, then reduce the power by one. A constant on its own differentiates to 0, and \(kx\) to \(k\). Before you differentiate, rewrite every root and every \(x\) in a denominator as a power: \(\sqrt{x} = x^{\frac{1}{2}}\), \(\dfrac{1}{x^{3}} = x^{-3}\), \(\dfrac{5}{\sqrt{x}} = 5x^{-\frac{1}{2}}\).
Answer: \(\dfrac{dy}{dx} = \dfrac{3}{\sqrt{x}} + \dfrac{16}{x^{3}}\); gradient at \(x = 4\) is \(\dfrac{7}{4}\)
Your turn
Find \(\dfrac{dy}{dx}\) when \(y = \dfrac{x^{2} + 3}{x}\).
Split first: \(y = x + 3x^{-1}\), so \(\dfrac{dy}{dx} = 1 - 3x^{-2} = 1 - \dfrac{3}{x^{2}}\).
Examiner trap
Never differentiate the top and bottom of a fraction separately: split \(\dfrac{x^{2} + 3}{x}\) into \(x + \dfrac{3}{x}\) first. And keep the constant where it belongs: \(\dfrac{1}{2x^{3}} = \tfrac{1}{2}x^{-3}\), not \((2x)^{-3}\).
Section 02
Six results to know by heart.
No calculus formulas are printed in the 0606 paper, so this table must be memorised. The trigonometric results are only true when \(x\) is in radians, so every calculus angle is in radians. Constant multiples and sums work exactly as with powers.
\(y\)
\(\dfrac{dy}{dx}\)
\(x^{n}\)
\(nx^{n-1}\)
\(\sin x\)
\(\cos x\)
\(\cos x\)
\(-\sin x\)
\(\tan x\)
\(\sec^{2} x\)
\(e^{x}\)
\(e^{x}\)
\(\ln x\)
\(\dfrac{1}{x}\)
Worked example First and second derivatives
Given \(y = 5\sin x - 3e^{x} + 2\ln x\), find \(\dfrac{dy}{dx}\) and \(\dfrac{d^{2}y}{dx^{2}}\).
Term by term: \(\dfrac{dy}{dx} = 5\cos x - 3e^{x} + \dfrac{2}{x}\).
Differentiate again, with \(\dfrac{2}{x} = 2x^{-1} \to -2x^{-2}\).
\(5\cos x \to -5\sin x\) and \(-3e^{x} \to -3e^{x}\).
Answer: \(\dfrac{dy}{dx} = 5\cos x - 3e^{x} + \dfrac{2}{x}\), \(\quad\dfrac{d^{2}y}{dx^{2}} = -5\sin x - 3e^{x} - \dfrac{2}{x^{2}}\)
Worked example Gradient at a point
Find the exact gradient of the curve \(y = 4\tan x - 2\cos x\) at the point where \(x = \dfrac{\pi}{4}\).
\(\dfrac{dy}{dx} = 4\sec^{2} x + 2\sin x\) (the minus in \(-2\cos x\) meets the minus in \(-\sin x\)).
At \(x = \dfrac{\pi}{4}\): \(\cos\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}\), so \(\sec^{2}\dfrac{\pi}{4} = 2\), and \(\sin\dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}\).
The sign of \(\cos x\): its derivative is \(-\sin x\), but the derivative of \(\sin x\) is \(+\cos x\). Also \(e^{x}\) does NOT follow the power rule (it is not \(xe^{x-1}\)), and a calculator in degree mode gives wrong gradients for trig functions.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
Find \(\dfrac{dy}{dx}\) when \(y = 5x^{4}\).
Quick check+10 XP
Find \(\dfrac{d}{dx}\left(\dfrac{2}{x}\right)\).
Quick check+10 XP
With \(x\) in radians, \(\dfrac{d}{dx}(\cos x) =\)
Quick check+10 XP
Find \(\dfrac{d}{dx}\left(e^{3x}\right)\).
Section 03
A function inside a function.
In \(y = (3x - 1)^{5}\) the bracket \(u = 3x - 1\) sits inside the outer function \(u^{5}\). The chain rule says \(\dfrac{dy}{dx} = \dfrac{dy}{du} \times \dfrac{du}{dx}\): differentiate the outer function leaving the inside alone, then multiply by the derivative of the inside. For a linear inside \(ax + b\) the extra factor is just \(a\).
\(\dfrac{d}{dx}(ax + b)^{n} = an(ax + b)^{n-1}\)
\(\dfrac{d}{dx}\sin(ax + b) = a\cos(ax + b)\) and \(\dfrac{d}{dx}\cos(ax + b) = -a\sin(ax + b)\)
\(\dfrac{d}{dx}\tan(ax + b) = a\sec^{2}(ax + b)\)
\(\dfrac{d}{dx}e^{ax + b} = ae^{ax + b}\)
\(\dfrac{d}{dx}\ln(ax + b) = \dfrac{a}{ax + b}\), and in general \(\dfrac{d}{dx}\ln f(x) = \dfrac{f'(x)}{f(x)}\)
Find the gradient of \(y = \dfrac{6}{(2x + 1)^{3}}\) at the point where \(x = 0\).
\(y = 6(2x + 1)^{-3}\), so \(\dfrac{dy}{dx} = 6 \times (-3)(2x + 1)^{-4} \times 2 = -\dfrac{36}{(2x + 1)^{4}}\). At \(x = 0\) the gradient is \(-36\).
Examiner trap
The most common lost mark is forgetting the inside derivative: \(\dfrac{d}{dx}\ln(5x + 2)\) is \(\dfrac{5}{5x + 2}\), not \(\dfrac{1}{5x + 2}\), and \(\dfrac{d}{dx}\sin 3x\) is \(3\cos 3x\), not \(\cos 3x\). Check by asking: what is inside, and did I multiply by its derivative?
Stage 2 · Exam
Exam-style questions
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Given \(y = (4x - 3)^{6}\), find \(\dfrac{dy}{dx}\).
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Find the gradient of the curve \(y = \ln(2x + 3)\) at the point where \(x = 1\).
Exam style+20 XP
Given \(y = 2\sin 3x\), find \(\dfrac{d^{2}y}{dx^{2}}\).
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Find the gradient of the curve \(y = \sqrt{x^{2} + 5}\) at the point where \(x = 2\).
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